Multiple ergodic averages with varying number of terms - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T02:49:15Z http://mathoverflow.net/feeds/question/64052 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64052/multiple-ergodic-averages-with-varying-number-of-terms Multiple ergodic averages with varying number of terms nonameisfinetoo 2011-05-05T20:52:11Z 2011-05-06T06:14:07Z <p>Hi. I've been stuck on the following question for some time.</p> <p>Consider a sequence of functions $\left( f_n \right)$ from an ergodic space $\left( \mathsf{X}, \mathsf{S}, \mu \right)$ to $\left[ 0,1 \right]$ such that $f_{a+b} \leq f_a \mathsf{S}^a \left( f_b \right)$ for all integers $a, b \geq 0$, where, classically, $\mathsf{S} \left( g \right)$ is the map $x \longmapsto g \left( \mathsf{S} \left( x \right) \right)$.</p> <p>Obviously $f_n \leq f_{n+1}$ so that the sequence $\left( f_n \right)$ decreases at each point to a function $f$. Under the hypothesis $\int_{\mathsf{X}} f_1 d\mu &lt; 1$ I was able to prove that $f = 0$ almost everywhere. Here's how I dealt with the problem:</p> <p>For fixed $n$ and integers $k$ and $\alpha$ such that $k \alpha \leq n$, the inequality $$f_n \leq f_1 \mathsf{S}^k \left( f_1 \right) \ldots \mathsf{S}^{k\alpha} \left( f_1 \right)$$ holds, so that, taking averages, one gets $$f_n \leq \frac{1}{\lfloor n/ \alpha \rfloor} \sum_{k=0}^{\lfloor n/ \alpha \rfloor} f_1 \mathsf{S}^k \left( f_1 \right) \ldots \mathsf{S}^{k\alpha} \left( f_1 \right)$$ Taking integrals, and using the dominated convergence theorem and the Furstenberg-Katznelson theorem on multiple ergodic averages, one gets $$\int_{\mathsf{X}} f \leq lim_{n \longrightarrow \infty} \frac{1}{\lfloor n/ \alpha \rfloor} \sum_{k=0}^{\lfloor n/ \alpha \rfloor} \int_{\mathsf{X}} f_1 \mathsf{S}^k \left( f_1 \right) \ldots \mathsf{S}^{k\alpha} \left( f_1 \right) = \left( \int_{\mathsf{X}} f_1 \right)^\alpha$$ For all integers $\alpha$, which allows me to conclude.</p> <p>Now the question is: does the series $\sum f_n$ converge? That seems plausible considering the seemingly exponential decreasing of $f_n$ to $f$, but trying to use the same techniques leads to multiple ergodic averages for a varying number of terms - making the integer $\alpha$ dependent on $n$ that is; is there any way to deal with those? All suggestions are welcome.</p> http://mathoverflow.net/questions/64052/multiple-ergodic-averages-with-varying-number-of-terms/64088#64088 Answer by R W for Multiple ergodic averages with varying number of terms R W 2011-05-06T06:14:07Z 2011-05-06T06:14:07Z <p>Unless I miss something, you are overcomplicating the story. In the first place, as it has already been pointed out, you don't really need submultiplicativity as $$f_n(x) \le f_1(x) f_1(Sx) \dots f_1(S^{n-1} x) = F_n(x) \;.$$ Applying the usual ergodic theorem to the right-hand side one gets that a.e. $$\log F_n(x)/n \to \int\log f_1(x)\;d\mu(x) &lt; 0 \;,$$ whence the claim.</p>